Question

以下是我打算合并的两个图：

首先是热图图的半矩阵。 ..............................

# plot 1 , heatmap plot
set.seed (123)
 myd <- data.frame ( matrix(sample (c(1, 0, -1), 500, replace = "T"), 50))

mmat <-  cor(myd)
diag(mmat) <- NA
mmat[upper.tri (mmat)] <- NA
heatmap (mmat, keep.dendro = F, Rowv = NA, Colv = NA)

enter image description here

我需要抑制x和y列中的名称并将它们放在对角线上。

第二个图，请注意第一个图中的名称/标签对应第二个图中的名称（x1到X10）：

  vard <- data.frame ( position = c(1, 10, 15, 18, 20, 23, 24, 30, 35, 40), 
          Names =paste ("X", 1:10, sep = ""))
    plot(vard$position, vard$position - vard$position,
                type = "n", axes = FALSE, xlab = "", ylab = NULL, yaxt = "n")
    polygon(c(0, max(vard$position + 0.08 * max(vard$position)),
                max(vard$position) + 0.08 * max(vard$position),
                0), 0.2 * c(-0.3, -0.3, 0.3, 0.3), col = "green4")
    segments(vard$position, -0.3, vard$position,                0.3)
    text(vard$position, 0.7, vard$position,
                    srt = 90)
    text(vard$position, -0.7, vard$Names)

enter image description here

我打算旋转第一个绘图，使X1到X10应该对应于第二个绘图中的相同，并且第二个绘图到第一个绘图中的标签之间存在连接。输出看起来像：

enter image description here 我怎样才能做到这一点？

编辑：根据有关add = TRUE的评论....我正在尝试将多边形添加到热图图中，如下所示。但是我找不到坐标。这种策略以这种方式绘制并稍后翻转实际图形......非常感谢......

enter image description here

Answer 1

这是一个完全基于网格的解决方案。唯一真正涉及的是函数convertToColors();它需要一个数字矩阵（可能包括NA）并将其转换为表示红色到白色"#FFFFFF比例颜色的sRGB颜色字符串（例如heat.colors()）。红色对应于矩阵中的最小值，白色对应于最大值，NA是透明的。

除此之外，我认为代码可以很好地显示有多少 grid 函数并不比低级 base <更复杂，并且更加一致和灵活/ strong>图形功能。

library(grid) ## Data: heatmap set.seed (123) myd <- data.frame ( matrix(sample (c(1, 0, -1), 500, replace = "T"), 50)) mmat <- cor(myd) diag(mmat) <- NA mmat[upper.tri (mmat)] <- NA ## Data: Positions vard <- c(1, 10, 15, 18, 20, 23, 24, 30, 35, 40) ## Construct a function to convert a numeric matrix to a matrix of color names. ## The lowest value in the matrix maps to red, the highest to white, ## and the NAs to "transparent". convertToColors <- function(mat) { # Produce 'normalized' version of matrix, with values ranging from 0 to 1 rng <- range(mat, na.rm = TRUE) m <- (mat - rng[1])/diff(rng) # Convert to a matrix of sRGB color strings m2 <- m; class(m2) <- "character" m2[!is.na(m2)] <- rgb(colorRamp(heat.colors(10))(m[!is.na(m)]), max = 255) m2[is.na(m2)] <- "transparent" return(m2) } ## Initialize plot and prepare two viewports grid.newpage() heatmapViewport <- viewport(height=1/sqrt(2), width=1/sqrt(2), angle = -135) annotationViewport <- viewport(y = 0.7, height = 0.4) ## Plot heat map pushViewport(heatmapViewport) grid.raster(t(convertToColors(mmat)), interpolate = FALSE) upViewport() ## Precompute x-locations of text and segment elements n <- nrow(mmat) v_x <- vard/max(vard) X_x <- seq(0, 1, len=n) ## Plot the annotated green bar and line segments pushViewport(annotationViewport) ## Green rectangle grid.polygon(x = c(0,0,1,1,0), y = c(.45,.55,.55,.45,.45), gp = gpar(fill = "green4")) pushViewport(viewport(width = (n-1)/n)) ## Segments and text marking vard values grid.segments(x0 = v_x, x1 = v_x, y0 = 0.3, y1 = 0.7) grid.text(label = vard, x = v_x, y = 0.75, rot = 90) ## Text marking heatmap column names (X1-X10) grid.text(paste0("X", seq_along(X_x)), x = X_x, y=0.05, gp = gpar(fontface="bold")) ## Angled lines grid.segments(x0 = v_x, x1 = X_x, y0 = 0.29, y1 = 0.09) upViewport() upViewport()

Answer 2

这不是一个完整的答案，但有一些想法可以帮助你构建一个......

与 base 图形系统相比，网格系统（ggplot2和lattice都基于此系统）多更好地支持在复合图中排列多个图形元素。它使用“视口”来指定图中的位置;任何高度，宽度和旋转度的视口都可以“推”到现有绘图中的任何位置。然后，一旦被推动，它们就可以被绘制成并最终升级，以便另一个地块可以放置在主绘图区域的其他地方。

如果这是我的项目，我可能会致力于一个完全基于网格的解决方案（自由使用更高级别的lattice或ggplot2图）。但是，gridBase包确实为组合 base 和 grid 图形提供了一些支持，我在下面的示例中使用了它。

（有关我在以下内容中所做的更多详细信息，请参阅grid.pdf中的viewports.pdf，rotated.pdf和file.path(.Library, "grid", "doc")个插图，以及通过键入vignette("gridBase", package="gridBase")）打开的小插图。

## Load required packages
library(lattice); library(grid); library(gridBase)

## Construct example dataset
set.seed (123)
myd <- data.frame ( matrix(sample (c(1, 0, -1), 500, replace = "T"), 50))
mmat <-  cor(myd)
diag(mmat) <- NA
mmat[upper.tri (mmat)] <- NA

## Reformat data for input to `lattice::levelplot()`
grid <- data.frame(expand.grid(x = rownames(mmat), y = colnames(mmat)), 
                   z = as.vector(mmat))

## Open a plotting device    
plot.new()     

## Push viewport that will contain the levelplot; plot it; up viewport.
pushViewport(viewport(y = 0.6, height = 0.8, width = 0.8, angle=135))
    lp <- levelplot(z~y*x, grid, colorkey=FALSE, 
                    col.regions=heat.colors(100), aspect=1,
                    scales = list(draw=FALSE), xlab="", ylab="", 
                    par.settings=list(axis.line=list(col="white")))
    plot(lp, newpage=FALSE)
upViewport()

## Push viewport that will contain the green bar; plot it; up viewport.
pushViewport(viewport(y = 0.7, height=0.2))
    # Use the gridBase::gridOMI to determine the location within the plot.
    # occupied by the current viewport, then set that location via par() call
    par(omi = gridOMI(), new=TRUE, mar = c(0,0,0,0))
    plot(0:1, 0:1,type = "n", axes = FALSE, xlab = "", ylab = "", yaxt = "n")
    polygon(x=c(0,0,1,1,0), y = c(.4,.6,.6,.4,.4), col = "green4")
upViewport()

enter image description here

Answer 3

我在这行代码中遇到错误：

myd <- data.frame ( matrix(sample (c(1, 0, -1), 500, replace = "T"), 50))

用TRUE（无引号）替换“T”来解决

结合r中的两个图

3 个答案: